Not if the operator is hermitian. Otherwise it can be. For instance, think about the situation where ##\Psi## is normalized to unity and the operator is just a multiplication with ##i## (imaginary unit).

Not if the operator is hermitian. Otherwise it can be. For instance, think about the situation where ##\Psi## is normalized to unity and the operator is just a multiplication with ##i## (imaginary unit).

Thank you very much! I have known the deduction.
According to the definition of Hermitian operator denoted by ##\hat{O}## as follows, $$<f|\hat{O}g>=<\hat{O}f|g>,$$ and making ##f=g##, we have $$<f|\hat{O}f>=<\hat{O}f|f>=(<f|\hat{O}f>)^{*},$$ meaning ##<f|\hat{O}f>## is real, which resembles the equation $$a=a^{*},$$ meaning ##a## is real.